# Uniqueness of Neumann series

Let $$f$$ be an entire function. Then there exist numbers $$a_0,a_1,\ldots$$, independent of $$z$$, such that $$f(z)=\sum_{n=0}^\infty a_n J_n(z),\quad \forall z\in\mathbb{C}$$ where $$J_n$$ is the Bessel function of the first kind of order $$n$$. This is called a Neumann series of $$f$$.

Question

Given $$f$$, are the numbers $$a_n$$ unique?

Thoughts

I took a look at Abramowitz and Stegun and Whittaker and Watson; in each book they mention the series but say nothing about the uniqueness of the coefficients. It would be nice if Neumann series turn out to be unique just as Taylor series of entire functions. I figured out that it is sufficient to prove that, for all $$z\in\mathbb{C}$$,

$$\sum_{n=0}^\infty a_n J_n(z)=0\implies \forall n\in\mathbb{N}:a_n=0;$$

Whittaker and Watson arrived at one possible integral representation of $$a_n$$ in terms of "Neumann polynomials". They essentially proved that if $$a_n=\text{integral expression}$$ then $$f(z)=\sum_{n=0}^\infty a_n J_n(z)$$, not the other way around.

This question has also been asked on MSE as can be seen at this link. According to the user Conrad in the comments, the uniqueness is known if the Neumann series are required to be uniformly convergent, but not known if they're only pointwise convergent.

Edit:

I'm thinking about using "orthogonality", i.e. $$\int_0^\infty \frac{J_{v+2l+1}(t)J_{v+2m+1}(t)}{t}\, dt=\frac{\delta_{l,m}}{2(2l+v+1)}$$ where $$v+l+m\gt -1$$ and then concluding that if $$\sum_{n=0}^\infty a_n J_n (z)=\sum_{n=0}^\infty b_n J_n(z),$$ then $$a_n=b_n$$ by the above orthogonality identity, but this requires me to justify switching $$\sum_{n=0}^\infty$$ with $$\int_0^\infty.$$

Please note that nothing about uniform convergence is assumed in this question; suppose that Neumann series only converge pointwise, nothing more.

• As a side note, on the whole real line, the orthogonal relation for the $\{J_n\}_{n\in\mathbb N}$ is somehow more complete because here $\int_{\mathbb R}J_n(t)J_m(t)\frac1{|t|}dt=0$ also holds for every odd $n$ and even $m$, since $J_n$ is an odd function and $J_m$ is an even function. Commented Sep 4 at 16:14
• @Yemon even in a Hilbert space, pointwise convergence means nothing per se - the trigonometric series case shows - if $0=\sum c_ne^{inx}$ with pointwise convergence for all $x$, then yes we can say that $c_n=0$ but this is a highly nontrivial result (and one that doesn't hold if we just have a.e. convergence pointwise); if we have $\sum c_ne^{inx}$ converges in $L^2$ (and hence ae) then $c_n=0$ is a trivial result; here, on the other hand, pointwise convergence gives local uniform convergence on a large open set by analyticity and that is why I believe the solution is unique Commented Sep 4 at 22:30
• (also, $J_k$ is not in the Hilbert space I mentioned for $k\le 0$) Commented Sep 4 at 22:32
• @Conrad Indeed. I was just trying to point out why a lot of the discussion by other people here might be misdirected. People seem to be pointing out (basic) facts about orthogonal systems in Hilbert spaces without looking at the setting of the original question. (And yes, I am aware of Montel's theorem and related ideas) Commented Sep 4 at 22:33
• Is this not evident from the fact that the Taylor expansion of $J_n(z)$ has lowest degree $n$? So (by induction) in order to make the $n$-th derivative of the sum vanish at $z=0$ the coefficient $a_n$ has to be zero too. Commented Sep 5 at 7:45

Let me show that the convergence is uniform on balls, so that one can differentiate term by term and find the $$a_n$$ by evaluating at $$z=0$$, using that $$J_n=c_nz^n+\dots$$. It is more convenient to use modified Bessel functions $$I_n(z)=(-i)^nJ_n(iz)=\sum_{m=0}^\infty\frac{1}{\Gamma(m+n+1)m!}\left (\frac{z}{2} \right)^{n+2m}.$$ Then $$f(iz)=\sum_{n} a_n J_n (iz)=\sum_n a_n i^n I_n(z)$$.

Note that $$I_n (1) \geq \frac{1}{n!2^n}$$ and that $$|I_n(z)| \leq \frac{1}{2^n n!}|z|^n e^{|z|^2/4}$$.

Since the series converges at $$x=1$$ we get $$|a_n| \leq Cn! 2^n$$ and this gives uniform convergence on any ball or radius less than 1.

• Great, you beat me on this. I was about to show from the series expansion that for $x<2$ one has $J_{n+1}(x)/J_n(x) < \frac{x}{2\,n}$, which also implies uniform convergence on sufficiently small intervals $[0,\epsilon]$. Commented Sep 6 at 18:36
• Awesome! You're getting the bounty. Commented Sep 6 at 19:37
• uh I didn't notice this answer, even more synthetic Commented Sep 6 at 21:33

edit: This was meant to recall a first elementary but relevant fact that was not mentioned at all, that is orthogonality.

Bessel functions $$\{J_n\}_{n\in\mathbb N_+}$$ are orthogonal on $$\mathbb R$$ w.r.to the measure $$dx/|x|$$. So existence and uniqueness of the expansion, in the corresponding Hilbert space, is just an instance of the well-known general fact [edit: that is the Fourier expansion of an element of a Hilbert space in a given orthonormal basis, here the space spanned by the $$J_n$$'s].

 in fact, since the degree of the lowest degree non-zero term of $$J_n$$ is $$n$$, the series makes sense even at the level of formal power series (for every sequence of coefficients), and you have uniqueness there (i.e. in the broader sense), and existence ($$a_n$$ are determined inductively from an infinite triangular system).

• To use orthogonality, you need a Hilbert space to which every entire function belongs. What is this Hilbert space? Commented Sep 4 at 11:36
• Actually I was thinking that using Osgood theorem (pointwise convergence implies locally uniform convergent on a dense open set in the holomorphic case) may work though of course $0$ may not be in the open dense set from Osgood so one would need to use a circle not centered there where the convergence is uniform and integrate term by term against some related functions to $O_n$ and get $a_n$ that way Commented Sep 4 at 12:48
• I'm still confused about something. Clearly the space in which the author's question is posed is not contained in $L^2({\mathbb R}, dx/|x|)$ since one is supposed to be able to represent any entire function as a pointwise-convergent infinite combination of the $J_n$. Perhaps the issue is that the reduction of the original "Question" to the statement about "if the sum of the series is zero, are all the terms zero" requires more justification? Commented Sep 4 at 18:10
• @Pietro Majer: This Hilbert space certainly does not contain ALL entire functions. Commented Sep 5 at 11:45
• Consider a similar but simpler problem: every entire function can be expanded into a series of exponentials $e^{\lambda_k z}$ with some complex lambdas. But this expansion is not unique. Commented Sep 5 at 11:47

We can say a bit more: if a Neumann series converges point-wise, it also converges absolutely and in $$\mathbb C[[z]]$$, and the limit is an entire function.

Let $$(a_n)_{n\ge0}$$ be a sequence of complex numbers such that the Neumann series $$\sum_{n=0}^\infty a_n J_n(z)$$ converges point-wise on $$\mathbb C$$. Let $$z\in\mathbb C$$ and $$r:=|z|$$. We have (by discrete Tonelli) $$\sum_{k\ge0,\, n\ge0} \Big|a_n \frac{(-1)^k}{k!(n+k)!}\big(\frac z2\big)^{n+2k}\Big|=\sum_{n\ge0}|a_n| \sum_{k\ge0}\frac1{k!(n+k)!}\big(\frac r2\big)^{n+2k}.$$ The inner sum in $$k$$ writes $$\sum_{k\ge0}\frac1{k!(n+k)!}\big(\frac r2\big)^{n+2k}= i^{-n}\sum_{k\ge0}\frac{(-1)^k}{k!(n+k)!}\big(\frac {ir}2\big)^{n+2k}=|J_n(ir)|\le\frac{|J_n(2ir)|}{2^n}.$$ By assumption, the series $$\sum_{n=0}^\infty a_n J_n(2ir)$$ is also convergent, so $$\big|a_n J_n(2ir)\big|\le M$$ for some constant $$M$$. Then the above the double sum is majorized by $$\sum_{n=0}^\infty\big|a_n J_n(2ir)\big|\frac1{2^n}\le \sum_{n=0}^\infty\frac M{2^n}=2M<+\infty.$$

So the family of terms $$\big\{a_n \frac{(-1)^k}{k!(n+k)!}\big(\frac z2\big)^{n+2k}: (n,k)\in\mathbb N^2\big\}$$ is absolutely summable. Thus (by discrete Fubini) we can change order of summation and write $$\sum_{n=0}^\infty a_n J_n(z)$$ as a power series of an entire function $$f(z)$$; moreover, as already observed, $$\sum_{n=0}^\infty a_n J_n(z)$$ also converges in the sense of formal power series, and $$a_n$$ are inductively determined by the coefficients of $$f$$ .

• Thank you! Your answer helped me a lot as well. If only I could split the bounty between two people... Commented Sep 7 at 0:25
• The first equation is suspicious, should it be inequality? Commented Sep 7 at 4:56
• The double sum has an equality, because everything is non-negative Commented Sep 7 at 6:15
• @Nomas2 It's fine, you did the right thing, and we are sportmen. Also thank you for the interesting question Commented Sep 7 at 10:33